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**enter image description here So as we can see the pulley attached to the body here is a movable pulley.In Illustration (A) if the block attached to the pulley is moved in the direction of the arrow with an displacement of X meters then we can state that the pulley attached to the object via string will also move with the block by a displacement of X meters.Now,if the pulley displaces by X meters then to keep the string around the pulley tight,the string has to move by 2x meters(for this problem assume the string movement is from bottom string to top string).Now coming to the second case i.e (B),here again everything is same as it was in (A),except the bottom string.In case (A) the bottom string was parallel to horizontal(I have forgot to draw it) where as in case (B) it is making and angle theta with the horizontal.

So my question is,in case (B)if the block will displace X meters along the direction of the arrow,the pulley will also move with it,so will this time the displacement of the strings(from bottom to top) to keep the string tight around the pulley be 2x meters.

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  • $\begingroup$ (a) This is a geometrical question rather than a physical. (b) Assuming that the top left hand end of the string is anchored, then X m of string moves upwards over the pulley in both cases. (c) But we really do need to know where the left hand ends of the string go. How, for example is angle $\theta$ maintained? $\endgroup$
    – Philip Wood
    Sep 6, 2020 at 14:57
  • $\begingroup$ Yes sir,the angle is maintained.Also you can assume that the bottom end of the string is attached to a small block nd the top end is attached to a rigid wall,in both case $\endgroup$
    – PATRICK
    Sep 6, 2020 at 15:07
  • $\begingroup$ Are you assuming that the string is of fixed length, and that the 'small block' attached to the bottom end of the string is moveable? In that case you need to draw 'before and after' diagrams for the two cases. These should give you the answers you want. $\endgroup$
    – Philip Wood
    Sep 6, 2020 at 15:27

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From the figure, there appear to be two possible cases: the height from the ground is maintained, or the angle is maintained. Constraint equations will be useful in solving both these cases.

If the angle is maintained, then the object will move $X(1 + \cos \theta)$ in the direction of the string.

If the height is maintained, then the object will move ${X(1 + \sec \theta)}$ horizontally, along the floor.

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  • $\begingroup$ No sir, that's not the case,in the case(B) the string will move x units as that of the red block attached to the movable pulley. Because if we imagine the string to get pulled x units away with the red block then assuming the string to be tight around the pulley,the upper string will move x units in order to stay attached to the pulley,but the below string will not move.Hence after X units of movement of the red block,the below string will move in an anticlockwise manner around the movable pulley with and displacement of X units. $\endgroup$
    – PATRICK
    Sep 8, 2020 at 12:55
  • $\begingroup$ I think you are confused, and I'm pretty sure I got it right. Read about constraint equations, you can solve it using that. Also, by height is fixed, I meant the block stays on the floor, and moves parallel to it. $\endgroup$
    – dnaik
    Sep 8, 2020 at 13:04
  • $\begingroup$ You might be right sir.Is there anyway I can contact you if you don't mind,I have a doubt. $\endgroup$
    – PATRICK
    Sep 9, 2020 at 15:30
  • $\begingroup$ You can ask the doubt here itself in the comments, its your own question, nobody will mind you clogging up the comments. I'll answer your doubt as a new answer or as an edit to my answer, depending on what it is. $\endgroup$
    – dnaik
    Sep 9, 2020 at 16:25
  • $\begingroup$ Right sir,but,In the comments I cannot paste pictures,to express my doubts to you,and it's time consuming,to draw everything on MSpaint and then post it here,because of which it would had been easier for me if we could have connected by an social media,but it totally depends on ypu sir...... Thank you for your time $\endgroup$
    – PATRICK
    Sep 10, 2020 at 15:23

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